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Sagot :

Answer:

r = [tex]\frac{1}{2}[/tex] [tex]\sqrt{l^2+4h^2}[/tex]

Step-by-step explanation:

h bisects l at right angles.

There is a right triangle formed with legs h and [tex]\frac{1}{2}[/tex] l and hypotenuse r

Using Pythagoras' identity, then

r² = ([tex]\frac{1}{2}[/tex] l )² + h²

   = [tex]\frac{1}{4}[/tex] l² + h² ← factor out [tex]\frac{1}{4}[/tex]

    = [tex]\frac{1}{4}[/tex] (l² + 4h²)

Take the square root of both sides

r = [tex]\sqrt{\frac{1}{4}(l^2+4h^2) }[/tex] = [tex]\frac{1}{2}[/tex] [tex]\sqrt{l^2+4h^2}[/tex]

Mhanifaidn

Answer:

  • r = [tex]\sqrt{(l/2)^2 + h^2}[/tex]

Step-by-step explanation:

Connect the endpoints of the with the center. The formed triangle is isosceles with sides l, r and r.

The height of the triangle h, is the distance between the center of the circle and the midpoint of the chord.

As per Pythagorean theorem we have:

  • (l/2)² + h² = r²
  • [tex]\sqrt{(l/2)^2 + h^2}[/tex] = r
  • r = [tex]\sqrt{(l/2)^2 + h^2}[/tex]

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